Question

The molar specific heat at constant pressure of
an ideal gas is (7/2) R. The ratio of specific heat at
constant pressure to that at constant volume is
(a) 8/7 (b) 5/7
(c) 9/7 (d) 7/5

Answer 1

Answer:

ans is c))9/7 hope help u mate ✌✌✌✌✌

Answer 2

$</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1</p><p></p><p>Solve the following by both substitution and elimination methods.</p><p></p><p>2x - \sqrt{2} y = 02x− </p><p>2</p><p> </p><p> y=0</p><p></p><p>and</p><p></p><p>\frac{3x}{ \sqrt{2} } - y = 1 </p><p>2</p><p> </p><p> </p><p>3x</p><p> </p><p> −y=1$