Question

A sample of air weighing 1.18 g occupies 1.0 × 103 cm3 when kept at 300 K and 1.0 × 105 Pa. When 2.0 cal of heat is added to it at constant volume, its temperature increases by 1°C. Calculate the amount of heat needed to increase the temperature of air by 1°C at constant pressure if the mechanical equivalent of heat is 4.2 × 107 erg cal−1. Assume that air behaves as an ideal gas.

Answer 1

The amount of heat needed to increase the temperature of air by 1°C at constant pressure if the mechanical equivalent of heat is 2.08 cal.

Explanation:

Step 1:

$m=1.18 g=1.18 \times 10^{-3} \mathrm{kg}$

ΔQ = 2.0×4.2 J

${P=1.0 \times 10^{6} \mathrm{Pa}} \\{V=1.0 \times 10^{3} \mathrm{cm} 3=1.0 \times 10^{-3} \mathrm{m} 3}\end{array}$

T = 300 K

Step 2:

Applying eqn. of state

PV = nRT

$n=\frac{P V}{R T}$

$n=1.0 \times 105 \times 1.0 \times 10-\frac{3}{8.314 \times 300}$

n = 0.04

Step 3:

ΔT = 10 C

$\Delta H_{v}=n C_{v} \Delta T$

$2.0 \times 4.2=n C_{v} \times 1$

$C_{v}=\frac{8.4}{n}=\frac{8.4}{0.04}$

$C_v$  = 210

Step 4:

Again we know

$c_{p}-C_{v}=R$

$c_{p}=R+C_{v}$

$c_{p}=8.3+210$

$c_{p}=218.3$

Step 5:

Now at constant pressure

$\Delta H_{p}=n c_{p} \Delta T$

$\Delta H_{p}=218.3 \times 0.04 \times 1=8.732 J$

In calories

$\Delta H_{p}=\frac{8.732}{4.2}=2.08 \mathrm{cal}$

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

2.08 cal

will be the answer

hope it help