Question

13. 2 moles of an ideal gas at temperature 27°C is
heated isothermally from volume V to 4V. If R = 2
cal/mol K, then the heat input in this process is
approximately
(1) 1560 cal
(2) 1660 cal
(3) 1760 cal
(4) 1860 cal​

Answer 1

2 1660 is the answer........

Answer 2

Answer:

(2) 1660 cal

Explanation:

We are given that

Number of moles=2

Temperature=$27^{\circ}=273+27=300 K$

R=2 cal/mol K

Volume increases from V to 4V

At isothermal (temperature constant)

We have to find the heat input in thi process

$\Delta Q=nRT ln\frac{V_2}{V_1}</p><p>Substitute the values then we get </p><p>[tex]\Delta Q=2\times 2\times 300ln{\frac{4V}{V}$

$\Delta Q=1200ln(2^2}$

$\Delta Q=2\times 1200 ln2=1663.5 cal$

$\Delta Q=1660 (Approximately)$

Answer:(2) 1660 cal